Question

Find the distance between these points.

A (5, 8), B(-3, 4)

I got AB = 4√5

is this correct?

Answer 1

yes \( \bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&(5\quad ,&8)\quad &(-3\quad ,&4)
\end{array}\qquad
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\ \quad \\
\implies d = \sqrt{(-3-5)^2 + (4-8)^2}\implies d=\sqrt{(-8)^2+(-4)^2}\\ \quad \\
d=\sqrt{64+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}\)

Answer 2

Thank you!

Answer 3

I'm not sure why I'm having an issue with this one but it's not coming out right when I do it. Would you mind helping?

Find the distance between these points.

W(-6, -8), X(6, 8)

Answer 4

I did d = \[\sqrt{(-6-6)^{2}+ (-8-8)^{2}}\]
d=\[\sqrt{(-12)^{2} + (-16^{2}}\]
d=\[\sqrt{144 + 256}\]
d=400

Answer 5

\(\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&(-6\quad ,&-8)\quad &(6\quad ,&8)
\end{array}\qquad
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\ \quad \\
\implies d = \sqrt{(6-(-6))^2 + (8-(-8))^2}\implies d=\sqrt{(12)^2+(16)^2}\\ \quad \\
d=\sqrt{144+256}\implies d=\sqrt{400}\implies d=20\)

Answer 6

Oh, duh! I forgot to find the square root. Blonde moment. Thank you so much! You're awesome!

Answer 7

yw